Let A be a matrix of order 2 x 2, whose entries are from the set {0, 1, 2, 3, 4, 5}. If the sum of a

Q: Wait, why do we use Stars and Bars here? Isn't that for distributing identical items?

That is a great observation! The Stars and Bars method is indeed used for distributing identical items into distinct bins, which is mathematically equivalent to finding the number of non-negative integer solutions to the equation $x_1 + x_2 + x_3 + x_4 = n$. Here, each variable $a, b, c, d$ acts as a bin, and the sum $p$ acts as the identical items being distributed. It is a perfect fit for this problem.

Q: Why do we only subtract cases for $p=7$? What about $p=3$ and $p=5$?

Excellent question. The constraint is that each entry must be $\le 5$. For $p=3$ and $p=5$, the sum is less than or equal to $5$. If the sum of four non-negative integers is $3$ or $5$, it is physically impossible for any single integer to be $6$ or greater. Thus, all solutions are valid. For $p=7$, however, it is possible to have an entry like $6$ or $7$, which violates our constraint, so we must filter those out.

Q: How do I know I've found all the invalid cases for $p=7$?

To be systematic, look at the maximum possible value an entry can take. Since the sum is $7$, an entry can be $7$ or $6$. If an entry is $7$, the others must be $0$. If an entry is $6$, the remaining sum is $1$, which must be distributed among the other three variables. By breaking it down into these specific scenarios—one entry is $7$ or one entry is $6$—you ensure you cover all possibilities where the constraint is violated.

Question

Let $A$ be a matrix of order $2 	imes 2$, whose entries are from the set $\{0, 1, 2, 3, 4, 5\}$. If the sum of all the entries of $A$ is a prime number $p, 2 < p < 8$, then the number of such matrices $A$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, where each entry $a, b, c, d \in \{0, 1, 2, 3, 4, 5\}$. We are tasked with finding the number of such matrices where the sum of the entries equals a prime number $p$, such that $2 < p < 8$.

The prime numbers satisfying the condition $2 < p < 8$ are $p = 3$, $p = 5$, and $p = 7$. We must evaluate the number of non-negative integer solutions for each case, ensuring that no individual entry exceeds the constraint of $5$.

### The Combinatorial Tool: Stars and Bars

To find the number of non-negative integer solutions to the equation $a + b + c + d = p$, we utilize the Stars and Bars theorem. The number of solutions is given by the binomial coefficient:

$$\binom{n+r-1}{r-1}$$

In this problem, we have $r = 4$ variables. Thus, the formula simplifies to $\binom{p+3}{3}$.

### Case 1: The Sum is $p = 3$

For $p = 3$, the number of non-negative integer solutions is:

$$\binom{3+3}{3} = \binom{6}{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$$

Since the maximum possible value for any single entry is $3$, which is well within our constraint of $\le 5$, all **20** solutions are valid.

### Case 2: The Sum is $p = 5$

For $p = 5$, the number of non-negative integer solutions is:

$$\binom{5+3}{3} = \binom{8}{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$$

Because the total sum is $5$, it is impossible for any single entry to exceed $5$. Therefore, all **56** solutions are valid.

### Case 3: The Sum is $p = 7$

For $p = 7$, the total number of non-negative solutions without constraints is:

$$\binom{7+3}{3} = \binom{10}{3} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$$

However, we must subtract cases where an entry exceeds $5$. An entry can be $6$ or $7$.

If one entry is $7$, the set of entries is $\{7, 0, 0, 0\}$. The number of permutations is:

$$\frac{4!}{3!} = 4$$

If one entry is $6$, the remaining sum is $1$, which must be assigned to one of the other three variables. The set of entries is $\{6, 1, 0, 0\}$, and the number of permutations is:

$$\frac{4!}{2!} = 12$$

The total number of invalid cases is $4 + 12 = 16$. Thus, the valid solutions for $p = 7$ are $120 - 16 = 104$.

### Final Calculation

To find the total number of such matrices, we sum the valid solutions from each case:

$$20 + 56 + 104 = 180$$

The total number of matrices satisfying the given conditions is **180**.

Animated Solution for Mathematics - Permutations and Combinations: Let $A$ be a matrix of order $2 \times 2$ , whose entries are from the set ${0, 1, 2, 3, 4, 5}$ . If the sum of all the entries of $A$ is a prime number $p, 2 < p < 8$ , then the number of such matrices $A$ is

Enter Numerical Value:

Visualized Solution

Defining the Matrix $A$

The Sum Condition

Identifying Possible Primes

Case 1: Sum $p = 3$

Calculating Solutions for $p = 3$

Case 2: Sum $p = 5$

Case 3: Sum $p = 7$ (Total Solutions)

The Constraint Trap for $p = 7$

Identifying Invalid Solutions

Valid Solutions for $p = 7$

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Combinatorial Tool

Case 1

Case 2

Case 3

Final Calculation

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Enter Numerical Value:

Visualized Solution

Defining the Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The Sum Condition

Identifying Possible Primes

Case 1: Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=3

Calculating Solutions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=3

Case 2: Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=5

Case 3: Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=7 (Total Solutions)

The Constraint Trap for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=7

Identifying Invalid Solutions

Valid Solutions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=7

Final Calculation

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